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I think it's about time you finaly do some reading of oracle documentation, in particular The Concepts Manual.

Untill you don't have at least a slightest idea what each of your two queries are doing and untill you don't know some of the very basic Oracle database vocabulary, any further explanation will only add more confusion. But nevertheless, I'll try to address some of your questions:

First of all which is which? I presume the first statement means "how much space is needed to store all those rows" and the second means "the other is how much space is actually allocated by that table".

I'll leave this for your homework - I hope you'll managed to find out which of your two queries is doing what.

What does "all the rows" mean here?

It means all the rows that are currently in your PRODUCT table. It means NUM_ROWS from your first query.

And also what does "how much space is needed..." mean?

It means how much space do you need in the table to store those NUM_ROWS rows in it. Your query reported that you need approximately 14 Kb free space to store those rows in your table. So, if your create a new table to be 14Kb in size that would probably be sufficient to insert those rows in it. If you create your new table to be 64Kb, you'll be fine too - when you insert those rows that you currently have in your PRODUCT table, they will occupy about 14KB of your new table, leaving 50Kb free space in it. But if you create your new table to be 10Kb large, without the ability for it to dinamicaly add new extents, you won't be able to insert all of the rows from your PRODUCT table in it, because they require 14Kb (according to your formula), but you only have 10Kb free space in it.

Thridly I didn't understand what you meant by "the other is how much space is actually allocated by that table". You mean how is actually used? If so "used out of what (how much)?"

As I said, take a look at the Concepts manual and read about "blocks" and "segments" and "extents", about "data files" and "tablespaces". It will all become so clear......

Jurij ModicASCII a stupid question, get a stupid ANSI
24 hours in a day .... 24 beer in a case .... coincidence?

OK. Let's make this simple. The lack of detail in this answer is intentional!

Consider this:

You insert 1,000,000 rows into a new table, causing it to increase in size so it can hold all the rows. At the end of this load process the table is 1M in size. It's safe to say that it takes 1M to store 1,000,000 rows.

You then decide to delete 999,999 rows, leaving a single row in the table. The table has not shrunk so it is still 1M in size, but obviously has lots of emty space in it. Can we use the same logic as before to make a conclusion? No. If we did we would be saying that it takes 1M to store 1 row. That's obviously nonsense.

So what?

Well, measuring the size of a table by summing all the blocks or bytes gives you the size of the table, not the size of the data it is holding.

Estimating the size of the data based on num_rows and avg_row_length gives you an idea of the size of the data, not the current size of the table holding it.

What next?

Decide what you actually want to know:

- The size of the table currently.
- The size of the data in the table.
- Something else.

Hmmm lots of details. Good, now atleast I am clear that I lack certain basic fundamentals. Will give it a reading... and get back if I still have some clarifications. Thanks for all your time and patience guys.

Disclaimer: Advice is provided to the best of my knowledge but no implicit or explicit warranties are provided. Since the advisor explicitly encourages testing any and all suggestions on a test non-production environment advisor should not held liable or responsible for any actions taken based on the given advice.

Just encountered this post and thought some of the people encountering it as well and that would like to calculate existing row and table sizes in order to calculate the entire sizing impact of new rows in the database can use a script I wrote. The script predicts the extra size needed for new rows inserted into a table with the table indexes and all refereing tables (with foreign keys) as well. Might be useful as it was to me...